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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBlitzer 8th EditionCh. 6 - Matrices and DeterminantsProblema 45
Capitolo 7, Problema 45

In Exercises 45–48, explain why the system of equations cannot be solved using Cramer's Rule. Then use Gaussian elimination to solve the system.
Image showing a system of three linear equations with variables x, y, and z arranged in matrix form.

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Step 1: Write the system of equations in matrix form as A * X = B, where A is the coefficient matrix, X is the variable vector, and B is the constants vector. Here, A = [[2, -3, 2], [2, 3, -2], [2, -9, 6]], X = [x, y, z]^T, and B = [4, 6, 2]^T.
Step 2: Calculate the determinant of the coefficient matrix A, denoted as det(A). If det(A) = 0, Cramer's Rule cannot be used because it requires a non-zero determinant to find a unique solution.
Step 3: Since det(A) = 0 (as you will find upon calculation), explain that the system either has infinitely many solutions or no solution, so Cramer's Rule is not applicable.
Step 4: Use Gaussian elimination to solve the system. Start by writing the augmented matrix [A | B]: [[2, -3, 2, 4], [2, 3, -2, 6], [2, -9, 6, 2]].
Step 5: Perform row operations to reduce the augmented matrix to row-echelon form, then use back substitution to find the values of x, y, and z or determine if the system is inconsistent or dependent.

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Cramer's Rule and Determinants

Cramer's Rule solves a system of linear equations using determinants, applicable only when the coefficient matrix has a nonzero determinant. If the determinant is zero, the system is either dependent or inconsistent, making Cramer's Rule unusable.
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Cramer's Rule - 2 Equations with 2 Unknowns

Gaussian Elimination

Gaussian elimination is a method to solve systems of linear equations by transforming the augmented matrix into row-echelon form using row operations. This process simplifies the system, allowing back-substitution to find the solution or determine if none exists.
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Solving Systems of Equations - Elimination

Consistency and Dependence of Systems

A system is consistent if it has at least one solution and inconsistent if it has none. Dependence occurs when equations are multiples or linear combinations of others, leading to infinite solutions or no unique solution, which affects the applicability of solution methods.
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Classifying Systems of Linear Equations
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Domanda del libro di testo

In Exercises 37 - 44, perform the indicated matrix operations given that A, B and C are defined as follows. If an operation is not defined, state the reason.

A=[40−3501],B=[51−2−2],C=[1−1−11]A=\(\begin{bmatrix}\)4 & 0\\ -3 & 5\\ 0 & 1\(\end{bmatrix}\),B=\(\begin{bmatrix}\)5 & 1\\ -2 & -2\(\end{bmatrix}\),C=\(\begin{bmatrix}\)1 & -1\\ -1 & 1\(\end{bmatrix}\)

A(BC)

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Domanda del libro di testo

In Exercises 46–51, evaluate each determinant.

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Domanda del libro di testo

In Exercises 45–48, explain why the system of equations cannot be solved using Cramer's Rule. Then use Gaussian elimination to solve the system.

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Domanda del libro di testo

In Exercises 43–44, (a) Write each linear system as a matrix equation in the form AX = B (b) Solve the system using the inverse that is given for the coefficient matrix.

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Domanda del libro di testo

In Exercises 37–44, use Cramer's Rule to solve each system. {x+2z=42y−z=52x+3y=13\(\begin{cases}\) x + 2z = 4 \\ 2y - z = 5 \\ 2x + 3y = 13 \(\end{cases}\)

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Domanda del libro di testo

Evaluate each determinant.

∣470−56032−4∣\(\begin{vmatrix}\) 4 & 7 & 0 \\ -5 & 6 & 0 \\ 3 & 2 & -4 \(\end{vmatrix}\)

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